Question 174373: An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $275.66 with a standard deviation of $78.11. (a) At the 5 percent level of significance, does this sample prove a violation of the guideline that the average patient should pay no more than $250 out-of-pocket? State your hypotheses and decision rule. (b) Is this a close decision?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $275.66 with a standard deviation of $78.11.
(a) At the 5 percent level of significance, does this sample prove a violation of the guideline that the average patient should pay no more than $250 out-of-pocket? State your hypotheses and decision rule.
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Ho: u = 250
Ha: u > 250
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Critical value for a one-tail test with alpha = 5%: z = 1.645
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Test statistic: z(275.66) = (275.66-250)/[78.11/Sqrt(25)] = 1.6456
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Conclusion: Since the TS is not in the reject interval, Fail to reject Ho.
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(b) Is this a close decision?
Yes, the test statistic is very close to the critical value.
p-value = P(z > 1.6456) = 0.05024, which is very close to the alpha value.
There are only 5.024% of test results that could have provided stronger
evidence for rejecting Ho.
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Cheers,
Stan H.
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